Решебник по алгебре 11 класс Никольский Параграф 8. Уравнения-следствия Задание 34

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\[\textbf{а)}\ \sqrt{x^{2} + 3} + \log_{2}x = 2 + \log_{2}x\]

\[x > 0.\]

\[\sqrt{x^{2} + 3} = 2\]

\[x^{2} + 3 = 4\]

\[x^{2} = 1\]

\[x = \pm 1.\]

\[x = - 1 - не\ подходит.\]

\[Ответ:x = 1.\]

\[\textbf{б)}\log_{2}\frac{12}{- 3 - x} = \log_{2}{(1 - x)}\]

\[1 - x > 0\]

\[x < 1.\]

\[\frac{12}{- 3 - x} = 1 - x;\ \ \ x \neq - 3\]

\[12 = (1 - x)( - 3 - x)\]

\[(x - 1)(x + 3) = 12\]

\[x^{2} - x + 3x - 3 - 12 = 0\]

\[x^{2} + 2x - 15 = 0\]

\[D_{1} = 1 + 15 = 16\]

\[x_{1} = - 1 + 4 =\]

\[= 3 > 1\ (не\ подходит);\]

\[x_{2} = - 1 - 4 = - 5.\]

\[Ответ:x = - 5.\]

\[\textbf{в)}\log_{\frac{1}{3}}\left( x^{4} - 17x^{2} + \log_{2}x \right) =\]

\[= \log_{\frac{1}{3}}{(19x^{2} + \log_{2}x)}\]

\[x > 1.\]

\[x^{4} - 17x^{2} + \log_{2}x =\]

\[= 19x^{2} + \log_{2}x\]

\[x^{4} - 36x^{2} = 0\]

\[x^{2}\left( x^{2} - 36 \right) = 0\]

\[x = 0 < 1 - не\ подходит;\ \ \]

\[x^{2} = 36\]

\[x = \pm 6.\]

\[x = - 6 < 1 - не\ подходит.\]

\[Ответ:x = 6.\]

\[\textbf{г)}\log_{3}\left( x^{2} - \sqrt{x^{2} - 4} + 1 \right) =\]

\[= \log_{3}\left( x^{3} - \sqrt{x^{2} - 4} - 6x + 1 \right)\]

\[x^{2} - \sqrt{x^{2} - 4} + 1 =\]

\[= x^{3} - \sqrt{x^{2} - 4} - 6x + 1\]

\[x^{3} - x^{2} - 6x = 0\]

\[x\left( x^{2} - x - 6 \right) = 0\]

\[x = 0;\]

\[x^{2} - x - 6 = 0\]

\[x_{1} + x_{2} = 1;\ \ x_{1} \cdot x_{2} = - 6\]

\[x_{1} = 3;\ \ x_{2} = - 2.\]

\[Проверка.\]

\[x = 0:\]

\[\sqrt{0 - 4} < 0\]

\[не\ корень.\]

\[x = 3:\]

\[9 - \sqrt{9 - 4} + 1 =\]

\[= 27 - \sqrt{9 - 4} - 18 + 1\]

\[10 - \sqrt{5} = 10 - \sqrt{5} > 0.\]

\[x = - 2:\]

\[4 - \sqrt{4 - 4} + 1 =\]

\[= - 8 - \sqrt{4 - 4} + 12 + 1\]

\[5 = 5 > 0.\]

\[Ответ:x = 3;x = - 2.\]